The internal and external radii are estimated to be 6 cm and 8 cm respectively, to the nearest whole number. The height of the cylinder is exactly 14 cm.
(a) Determine the exact values for internal and external radii which will give maximum volume of the material used.
(b) Calculate the maximum possible volume of the material used. Take the value of Pi to be 22/7
for a cylinder, max volume will occur when the max cross-section area is achieved.
If the radii are exactly 6 and 8, then the area is π(8^2-6^2) = 28π
If the accuracy is within ±x, then we have the area πa satisfying
a = π((8+x)^2-(6-x)^2) = 28π(x+1)
So, the only constraint is the allowable values of x. Since the radii were estimated to the nearest integer, x <= 0.5
So the radius 5.5 <= r <= 8.5
and the volume is 14π(8.5^2-5.5^2) = 588π cm^3
To determine the radii that will give the maximum volume of the material used, we can use derivatives.
(a) Let's define the internal radius as r1 and the external radius as r2.
The volume V of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height.
For the internal cylinder, V1 = πr1^2h
For the external cylinder, V2 = πr2^2h
The volume of material used is V2 - V1 = πr2^2h - πr1^2h = πh(r2^2 - r1^2)
To find the radii that maximize the volume, we need to find the values of r1 and r2 that maximize the expression for the volume of material used. To do this, we can take the derivative of the volume with respect to r2 and set it equal to zero.
d/d(r2)(πh(r2^2 - r1^2)) = 0
2πh(r2) = 0
Simplifying the equation, we get r2 = 0.
Since r2 represents the external radius of the cylinder, it cannot be zero. Therefore, there is no maximum value for the volume of material used. The external radius can be any positive value greater than the internal radius.
(b) Given that the internal radius is 6 cm (r1 = 6 cm), the maximum possible volume of the material used can be calculated using the formula V = πh(r2^2 - r1^2).
Substituting the values r1 = 6 cm, r2 ≈ 8 cm (to the nearest whole number), and h = 14 cm into the formula:
V = π(14)((8)^2 - (6)^2)
= π(14)(64 - 36)
= (22/7)(14)(28)
≈ 22(2)(4)
= 176 cm^3
Therefore, the maximum possible volume of the material used is approximately 176 cm^3.
To determine the values for the internal and external radii that will give the maximum volume of the material used, we need to use optimization techniques.
(a) Let's denote the internal radius as r1 and the external radius as r2. We want to find the values of r1 and r2 that maximize the volume of the material used, which is the volume of the cylinder.
The volume of a cylinder is given by the formula: V = π * (r2^2 - r1^2) * h
In this case, the height (h) is given as exactly 14 cm, so we can substitute that in the formula: V = π * (r2^2 - r1^2) * 14
To find the maximum volume, we need to find the critical points by taking the derivative of the volume equation with respect to either r1 or r2 and setting it equal to zero. Let's take the derivative with respect to r1:
dV/dr1 = π * (-2r1) * 14 = -28πr1
Setting dV/dr1 equal to zero and solving for r1:
-28πr1 = 0
r1 = 0
We can see that r1 = 0 is a critical point, but since it's not a valid radius for the cylinder, we'll ignore it.
Next, let's take the derivative with respect to r2:
dV/dr2 = π * (2r2) * 14 = 28πr2
Setting dV/dr2 equal to zero and solving for r2:
28πr2 = 0
r2 = 0
Similarly, r2 = 0 is a critical point but not a valid radius, so we'll ignore it as well.
Since the internal and external radii must be positive, we can conclude that the maximum volume will occur at the endpoints of the given range for the radii, which are 6 cm and 8 cm respectively. Therefore, the exact values for the internal and external radii that will give the maximum volume are:
Internal radius (r1) = 6 cm
External radius (r2) = 8 cm
(b) To calculate the maximum possible volume of the material used, we can substitute these values into the volume formula:
V = π * (r2^2 - r1^2) * h
V = π * (8^2 - 6^2) * 14
Calculating this expression:
V = π * (64 - 36) * 14
V = π * 28 * 14
V = 392π cm^3
Now, taking the value of Pi to be 22/7:
V = 392 * (22/7)
V = 1232 cm^3
Therefore, the maximum possible volume of the material used is 1232 cm^3.